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I’ve had to deal with a question like this on a test on friday and I just can’t get it out of my mind.

I was supposed to calculate the moment of inertia of a cylinder with an inhomogenous yet continuous mass density distribution. The density was supposed to “increase with the square of the radius,” starting at $m=0$ for $r=0$.

Now, moment of inertia aside, the interesting part of this task (I think) is finding an expression for the mass density. For the mass density of a cylinder for which we put a coordinate system at its center we can say that $$\rho(\vec{r})=\frac{m}{\pi R^2L}\Theta(R-r)\Theta(\frac{L}{2}-|z|).$$ Where $R$ is the radius of the cylinder and $L$ is its height. This leaves the mass density at $\rho(\vec{r})=0$ anywhere outside of the cylinder and inside it’s just $\rho(\vec{r})=\frac{m}{V}$, but how would I go about meeting this condition above? The volume is constant; it’s not changing as there are no chips of the cylinder flying off, so the only way to make the mass density increase with the square of the radius is for the mass itself to be a function of $r$.

So here’s where I’m stuck: I could obviously write the mass as a function of $r$ like so: $$m(\vec{r})=m_0 r^2,$$ but that would completely mess up my units in the end. Not only would the mass suddenly no longer be $kg$, but $kgm^2$ ‒ the mass density $\rho$ would change as well, as it would turn from $\frac{kg}{m^3}$ to $\frac{kg}{m}$!

I just don’t see an approach to this. Am I misinterpreting the task? Am I missing something crucial or am I making things too complicated for myself?

I appreciate the help.

[UPDATE]

I get what you’re trying to say and I’ve actually tried this before, too, but I got a weird result.

Taking your function as an example, I can integrate over the volume using cylindrical coordinates: $$\begin{eqnarray}\rho(\vec{r})&=&\int_{0}^{R}{dr'r'}\int_{0}^{2\pi}{d\varphi'}\int_{-\frac{L}{2}}^{\frac{L}{2}}{dz'}Cr'^2\\ &=&C2\pi L\int_{0}^{R}{dr'r'^3}\\ \rightarrow\rho(\vec{r})&=&\frac{C\pi LR^4}{2}\end{eqnarray}$$ Now if I solve this for $C$ to take a look at the units, I get $$\Rightarrow C=\frac{2\rho}{\pi LR^4}$$ and therefore $$[C]=\frac{\frac{kg}{m^3}}{m^5}=\frac{kg}{m^8}$$ and that’s kind of . . . wrong?

The density would then be $$[\rho]=[Cr^2]=\frac{kgm^2}{m^8}=\frac{kg}{m^6}$$ and that can’t be right. Where did I screw up?

If I take the result of my integral and equate that to the total mass, my density would look even weirder: $$\begin{eqnarray}\rho&=&\frac{m}{V}\\ &=&\frac{C\pi LR^4}{2\pi R^2L}\\ \rho&=&\frac{CR^2}{2}\end{eqnarray}$$ In units the density would, again, look like so: $$[\rho]=\Bigg[\frac{CR^2}{2}\Bigg]=\frac{kgm^2}{m^8}=\frac{kg}{m^6}.$$

So, somewhere in all of this I’m making a terrible mistake and I don’t see where. :/

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  • $\begingroup$ use integration. The mass will depend on (r^4)*L times a constant. a little dimensional analysis will show that the dimensions of the constant is mass/[length^5]. Thus it is correct. $\endgroup$
    – Lelouch
    Jul 25, 2016 at 6:58

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What you want is $\rho(\textbf{r})=Cr^2$, where $C$ is a constant. Find $C$ by integrating this function over cylinder volume, and equating the result to total mass $m$.

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  • $\begingroup$ Thanks for the answer, see update above please. I’m still going wrong somewhere . . . $\endgroup$
    – Sephi-
    Jul 25, 2016 at 8:37
  • $\begingroup$ Equate the integral to mass: $m=\int[~~]$, and not to $\rho$ $\endgroup$
    – Deep
    Jul 25, 2016 at 11:51
  • $\begingroup$ OH! I think I’ve got it now. By equating the integral of the density over the volume to the mass, I obviously get reasonable results … I should’ve been able to figure that out. So for my mass I get $m=\frac{\pi R^4LC}{2}$ and my density turns to $\rho(\vec{r})=\frac{2mr^2}{\pi R^4L}$. Thanks for your reply and the tips. $\endgroup$
    – Sephi-
    Jul 25, 2016 at 14:14

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